JBMO 1998 Question | You Should Try This Amazing Math Olympiad Algebra Problem

You Should Try This Amazing Math Olympiad Algebra Problem | Square Root of a Large Number

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[田村博志-zy]

We note that
25 = 5^2
1225 = 35^2
We try calculating 335^2 and obtain 335^2 = 112225.
And so on. No need to consider a large number.

[cubetoast]

Alternatively we can use the trick that square of numbers ending in five is 100*n*(n+1)+25 where n is digits from left leaving 5 at the right end. For eg- 345^2=100*34*35+25
so we can clearly see that product of two consecutive numbers is where 1 is 1997 times and same for 2.
Now there is another trick
sqrt(4489) is 67 and if we add another 48 is between then the we add a 6 in sqrt
sqrt(444889) is 667
x(x+1)=111....1111....2222 Here we can use quadratic formula to get Here 4 is 1997 times and 8 is 1996 times.
So x is where 6 is 1997 times which is ±333....3333....333 where 3 is 1997 times
Now we add 5 at last so the final answer is ±3333....3333...3335 where 3 is 1997 times.
we can use quadratic formula and a bit of guess work to find the solution.

[vaibhavsunak]

Very tricky question....explained nicely

[surekharajput]

So nicely done and nice method to solve above tricky question.

[ratilaljarsania]

It can be easily solved by an alternate method.
Square of any number ending with 5 an be given as x*(x+1)25. I.e. 35 Square will be 3*4=12 and will be 1225. Likewise it can be calculated for any number ending with 5. Here also it ends with 25.
Now we have to divide the remaining number after removing 25 at the end in two consecutive numbers whose product is that number. By observing 33*34=1122, 3333*3334=11112222 and so on so we will get answer by this way as 33...333 for 1997 times.

[BbNn]

Fantastic!!!...very nice problem and its solution!

[mrdarjeeling]

Sorry to say that nothing went to my head 😢